for(i=0 to n-1)
   if( binarysearch(i,n-1,1) + 1)
   count++
print count.
binarysearch(first,last,item)
  if(1 is there)
  return mid
  else
  return -1.
similarly we can go for coloumns.
o(nlogn)

On 7/5/10, divya jain <sweetdivya....@gmail.com> wrote:
> i think u need to visit every element atleast once to see if its 1 or 0, nd
> so update the count.
> so i dont think it will be possible in less than O(n2)
>
> On 5 July 2010 15:41, amit <amitjaspal...@gmail.com> wrote:
>
>>
>>
>> For a given matrix NxN having 0 or 1’s only. You have to find the
>> count of rows and columns where atleast one 1 occurs.
>>
>> e,g
>>
>> 0 0 0 0
>>
>> 1 0 0 1
>>
>> 1 0 0 1
>>
>> 1 1 0 1
>>
>> Row count having 1 atleast once: 3
>>
>> Col count having 1 atleast once: 3
>>
>> Any Solution less than O(n^2) will do....
>>
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